Problem 1-9: Power of a Current Source

Find the power of  Is_1 using circuit reduction methods.
Resistive Circuits - Circuit Reduction Method


Solution
 R_1 and  R_4 are parallel.  R_2 and  R_3 are also parallel. Therefore:

Resistive Circuits - Parallel Resistors Reduction

where  R_{14} = R_1 || R_4 = 5 \Omega and  R_{23} = R_2 || R_3 = 3 \Omega . Now,   R_{14} and  R_{23} are in series. Hence,
Resistive Circuits - Series Resistors Reduction


and  R_{1234} = R_{14} + R_{23} = 8 \Omega . Voltage across the current source, i.e.  V_{Is_1} can be found by the resistor:
 V_{Is_1} = Is_1 \times R_{1234} = 80 v . Thus,

 P_{Is_1} = - V_{Is_1} \times Is_1 = -800 , supplying.

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